c) Energy Method n k=2 n Fk. Vk + Σ Tk. Wk = k=2 The force due to pressure is: mk .ak Vk + Ik ak. Wk (eq. 1) Fp=(Pressure) x (Surface) = P₁ ×л- k=2 d² 0.038² = 500 × 103 xπ- = 566.7 N and F = -566.7, V = d = 2.9867, ap = 181.02 m/s² Then, eq.1 =>-566.77 x 2.9867+ T12 × W₂ = mp xap xvp • • Power required: T12 W20.45 x 181.02 x 2.98267 + 566.77 x 2.9867 = 1935.74 Watt Torque required: T12= 1935.74 83.776 = 23.1 N.m Problem 3 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. The hydraulic cylinder (DB) is designed to lift the platform so that the arm (ABC) rotates, about the pin A, with a constant angular speed of 0.1 rad/s CW. For the position shown when 0 = 20°; a) Draw clearly the vector loop (DAB). b) Find the velocity of extension of the hydraulic cylinder (DB) and its angular velocity. c) Find the velocity of center of gravity C. d) Using Energy Method, determine the force F required by the hydraulic cylinder (DB) to lift the platform. Neglect the masses of all members. 5 m 2.4 m B 0.5 m A 0.9 m

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
icon
Related questions
Question
Please I want a detailed explanation of the energy method rule placed in the first line and how these numbers appeared to us. I want a detailed explanation please urgent .
c) Energy Method
n
k=2
n
Fk. Vk + Σ Tk. Wk =
k=2
The force due to pressure is:
mk .ak Vk + Ik ak. Wk (eq. 1)
Fp=(Pressure) x (Surface) = P₁ ×л-
k=2
d²
0.038²
= 500 × 103 xπ-
= 566.7 N
and F = -566.7, V = d = 2.9867, ap = 181.02 m/s²
Then, eq.1 =>-566.77 x 2.9867+ T12 × W₂ = mp xap xvp
•
•
Power required:
T12 W20.45 x 181.02 x 2.98267 + 566.77 x 2.9867 = 1935.74 Watt
Torque required:
T12=
1935.74
83.776
= 23.1 N.m
Transcribed Image Text:c) Energy Method n k=2 n Fk. Vk + Σ Tk. Wk = k=2 The force due to pressure is: mk .ak Vk + Ik ak. Wk (eq. 1) Fp=(Pressure) x (Surface) = P₁ ×л- k=2 d² 0.038² = 500 × 103 xπ- = 566.7 N and F = -566.7, V = d = 2.9867, ap = 181.02 m/s² Then, eq.1 =>-566.77 x 2.9867+ T12 × W₂ = mp xap xvp • • Power required: T12 W20.45 x 181.02 x 2.98267 + 566.77 x 2.9867 = 1935.74 Watt Torque required: T12= 1935.74 83.776 = 23.1 N.m
Problem 3
The telescoping arm ABC is used to provide an elevated platform for construction
workers. The workers and the platform together have a mass of 200 kg and have a
combined center of gravity located directly above C. The hydraulic cylinder (DB) is
designed to lift the platform so that the arm (ABC) rotates, about the pin A, with a
constant angular speed of 0.1 rad/s CW. For the position shown when 0 = 20°;
a) Draw clearly the vector loop (DAB).
b) Find the velocity of extension of the hydraulic cylinder (DB) and its angular
velocity.
c) Find the velocity of center of gravity C.
d) Using Energy Method, determine the force F required by the hydraulic
cylinder (DB) to lift the platform.
Neglect the masses of all members.
5 m
2.4 m
B
0.5 m
A
0.9 m
Transcribed Image Text:Problem 3 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. The hydraulic cylinder (DB) is designed to lift the platform so that the arm (ABC) rotates, about the pin A, with a constant angular speed of 0.1 rad/s CW. For the position shown when 0 = 20°; a) Draw clearly the vector loop (DAB). b) Find the velocity of extension of the hydraulic cylinder (DB) and its angular velocity. c) Find the velocity of center of gravity C. d) Using Energy Method, determine the force F required by the hydraulic cylinder (DB) to lift the platform. Neglect the masses of all members. 5 m 2.4 m B 0.5 m A 0.9 m
Expert Solution
steps

Step by step

Solved in 2 steps with 9 images

Blurred answer
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY